# Maximum Path Sum — Dynamic Programming Algorithm

I came across this problem recently that required solving for the maximum-sum path in a triangle array.

To copy the above triangle array:

 75 95 64 17 47 82 18 35 87 10 20 04 82 47 65 19 01 23 75 03 34 88 02 77 73 07 63 67 99 65 04 28 06 16 70 92 41 41 26 56 83 40 80 70 33 41 48 72 33 47 32 37 16 94 29 53 71 44 65 25 43 91 52 97 51 14 70 11 33 28 77 73 17 78 39 68 17 57 91 71 52 38 17 14 91 43 58 50 27 29 48 63 66 04 68 89 53 67 30 73 16 69 87 40 31 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
view raw euler18.txt hosted with ❤ by GitHub

As can be seen, there are 15 levels to this tree (including the top most node). Therefore, there are 214 possible routes to scan in order to check for the maximum sum using brute force. As there are only 214 (16384) routes, it is possible to solve this problem by trying every route. However, doing the same using brute force on a triangle array of 100 levels would take several billion years to solve using a computer that checks through say, 1012 routes per second. A greedy algorithm might per-chance work for the particular 4-level example problem stated above, but will not always work, and in most cases won’t. For instance, for the 100-level problem:

 59 73 41 52 40 09 26 53 06 34 10 51 87 86 81 61 95 66 57 25 68 90 81 80 38 92 67 73 30 28 51 76 81 18 75 44 84 14 95 87 62 81 17 78 58 21 46 71 58 02 79 62 39 31 09 56 34 35 53 78 31 81 18 90 93 15 78 53 04 21 84 93 32 13 97 11 37 51 45 03 81 79 05 18 78 86 13 30 63 99 95 39 87 96 28 03 38 42 17 82 87 58 07 22 57 06 17 51 17 07 93 09 07 75 97 95 78 87 08 53 67 66 59 60 88 99 94 65 55 77 55 34 27 53 78 28 76 40 41 04 87 16 09 42 75 69 23 97 30 60 10 79 87 12 10 44 26 21 36 32 84 98 60 13 12 36 16 63 31 91 35 70 39 06 05 55 27 38 48 28 22 34 35 62 62 15 14 94 89 86 66 56 68 84 96 21 34 34 34 81 62 40 65 54 62 05 98 03 02 60 38 89 46 37 99 54 34 53 36 14 70 26 02 90 45 13 31 61 83 73 47 36 10 63 96 60 49 41 05 37 42 14 58 84 93 96 17 09 43 05 43 06 59 66 57 87 57 61 28 37 51 84 73 79 15 39 95 88 87 43 39 11 86 77 74 18 54 42 05 79 30 49 99 73 46 37 50 02 45 09 54 52 27 95 27 65 19 45 26 45 71 39 17 78 76 29 52 90 18 99 78 19 35 62 71 19 23 65 93 85 49 33 75 09 02 33 24 47 61 60 55 32 88 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87 83 31 03 93 70 81 47 95 77 44 29 68 39 51 56 59 63 07 25 70 07 77 43 53 64 03 94 42 95 39 18 01 66 21 16 97 20 50 90 16 70 10 95 69 29 06 25 61 41 26 15 59 63 35
view raw euler67.txt hosted with ❤ by GitHub

The Algorithm

Solving such a problem would require a powerful approach – and surely enough, there is an algorithm that solves the 100-level problem in a fraction of a second. Here’s a brief sketch of the algorithm:

You have such triangle:

   3
7 4
2 4 6
8 5 9 3


Let’s say you’re on the penultimate level 2 4 6 and you have to iterate over it.

From 2, you can go to either 8 or 5, so 8 is better (maximize you result by 3) so you calculate the first sum 8 + 2 = 10

From 4, you can go to either 5 or 9, so 9 is better (maximize you result by 4) so you calculate the second sum 9 + 4 = 13

From 6, you can go to either 9 or 3, so 9 is better again (maximize you result by 6) so you calculate the third sum 9 + 6 = 15

This is the end of first iteration and you got the line of sums 10 13 15.

Now you’ve got triangle of lower dimension:

      3
7    4
10   13    15


Keep going this way…

         3
20    19


…and you finally arrive at 23 as the answer.

The Code

Now for the Python code. I first store the 100-level triangle array in a text file, euler67.txt
I read the triangle array into Python and successively update the penultimate row and delete the last row according to the algorithm discussed above.

 # Read the problem matrix into a triangle array in python filename = 'euler67.txt' with open(filename, "r") as ins: array = [] for line in ins: array.append(line) # Convert the triangle arry entries into integers newArray = [] for i in array: j = i.split(' ') k = [int(n) for n in j] newArray.append(k) l = len(newArray) # Algorithm to calculate Maximum Path Sum for i in range(l-1): array1 = newArray[-1] array2 = newArray[-2] for j in range(len(array2)): array2[j] += max(array1[j], array1[j+1]) newArray.pop(-1) newArray[-1] = array2 print newArray[0][0]
view raw euler67.py hosted with ❤ by GitHub

This code is the key to solving problems 18 and 67 of Project Euler.
Problem 18
Ans: 1074
Problem 67
Ans: 7273

# Collatz Conjecture — What You Need to Know

Like many of my previous posts, this post too has something to do with a Project Euler problem. Here’s a sketch of the Colatz Conjecture.

The following iterative sequence is defined for the set of positive integers:

n → n/2 (n is even)
n → 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

So basically, it’s just this. Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has aptly been called oneness! But perhaps oneness has its pitfalls too…

If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1. Such a sequence might enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.

Question
It can be seen that the sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

HUGE HINT:

Histogram of stopping times for the numbers 1 to 100 million. Stopping time is on the x axis, frequency on the y axis.

Approach 1 (A naïve, but straigh forward method)

 # Longest Collatz Sequence under a million # Function listing collatz sequence for a number def collatz(n): "function listing collatz sequence for a positive integer" coll = [] coll.append(n) while n != 1: if n % 2 == 0: n = n/2 coll.append(n) else: n = 3*n + 1 coll.append(n) return coll longest = 0 j = 0 for i in xrange(1, 1000000): lencoll = len(collatz(i)) if lencoll > longest: longest = lencoll j = i print j
view raw euler14.py hosted with ❤ by GitHub

Approach 2 (Smart, quick method that uses dynamic programming with the help of dictionaries)

 collatz = {1:1} def Collatz(n): global collatz if not collatz.has_key(n): if n%2 == 0: collatz[n] = Collatz(n/2) + 1 else: collatz[n] = Collatz(3*n + 1) + 1 return collatz[n] for j in range(1000000,0,-1): Collatz(j) print collatz.keys()[collatz.values().index(max(collatz.values()))]
view raw euler14.py hosted with ❤ by GitHub

I couldn’t help appreciate the elegance of the second algorithm. It’ll be well worth perusing if you don’t get it at one go. [Hint: It keeps track of the number of terms of a particular sequence as values assigned to keys of a Python dictionary]

Ans: 837799

# Highly Divisible Triangular Number — Project Euler (Problem 12)

All n numbers are Triangle Numbers. They’re called so, because they can be represented in the form of a triangular grid of points where the first row contains a single element and each subsequent row contains one more element than the previous one.

Problem 12 of Project Euler asks for the first triangle number with more than 500 divisors.

These are the factors of the first seven triangle numbers:

1 = 1: 1
2 = 3: 1,3
3 = 6: 1,2,3,6
4 = 10: 1,2,5,10
∑5 = 15: 1,3,5,15
∑6 = 21: 1,3,7,21
∑7 = 28: 1,2,4,7,14,28

Here’s how I proceeded:

First Step: Find the smallest number with 500 divisors. Seems like a good starting point to begin our search.
Second Step: Starting at the number found in the previous step, search for the next triangle number. Check to see whether this number has 500+ divisors. If yes, this is the number we were looking for, else…
Third Step: Check n for which ∑n = triangle number found in the previous step
Fourth Step: Add (n+1) to the last triangle number found, to find the next triangle number. Check whether this number has 500+ divisors. If yes, this number is the answer. If not, repeat Fourth Step till the process terminates.

Now for the details:

The First Step isn’t exactly a piece of cake, but necessary to reduce computation time. I solved this with a bit of mental math. The main tool for the feat is the prime number decomposition theorem:

Every integer N is the product of powers of prime numbers

N = pαqβ· … · rγ
Where p, q, …, r are prime, while α, β, …, γ are positive integers. Such representation is unique up to the order of the prime factors.
If N is a power of a prime, N = pα, then it has α + 1 factors:
1, p, …, pα-1, pα
The total number of factors of N equals (α + 1)(β + 1) … (γ + 1)

500 = 2 x 2 x 5 x 5 x 5
So, the number in question should be of the form abq4r4s4 where a, b, q, r, s are primes that minimize abq4r4s4. This is satisfied by 7x11x24x34x54 = 62370000. This marks the end of the First Step which is where we start our search for our magic number.

The next 3 steps would need helper functions defined as below:

 from math import * # Function to calculate the number of divisors of integer n def divisors(n): limit = int(sqrt(n)) divisors_list = [] for i in range(1, limit+1, 1): if n % i == 0: divisors_list.append(i) if i != n/i: divisors_list.append(n/i) return len(divisors_list) # Function to check for triangle number def isTriangleNumber(n): a = int(sqrt(2*n)) return 0.5*a*(a+1) == n # Function to calculate the last term of the series adding up to the triangle number def lastTerm(n): if isTriangleNumber(n): return int(sqrt(2*n)) else: return None
view raw euler12functions.py hosted with ❤ by GitHub

As can be seen from the above code, the algorithm to calculate divisors of an integer is as follows:
1. Start by inputting a number n
2. Let an int variable limit = √n
3. Run a loop from i = 1 to  i = limit
3.1 if n is divisible by i
3.1.1 Add i to the list of divisors
3.1.2 if i and n/i are unequal, add n/i to the list too.
4. End

Finally, executing the 4 steps mentioned earlier can be done like so (the code took less than 2s to arrive at the answer):

 # First Step # First number 'check' to have 500 divisors check = 2**4 * 3**4 * 5**4 * 7 * 11 # Second Step # Starting from 'check', iterate sequentially checking for the next 'triangle' number while not isTriangleNumber(check): check += 1 # Third and Fourth Steps # Calculate the last term of the series ('seriesLastTerm') that adds up to the newly calculated triangle number 'check' seriesLastTerm = lastTerm(check) # Iterate over triangle numbers checking for divisors > 500 while divisors(check) <= 500: # add the next term to check to get the next triangle number check += (seriesLastTerm + 1) seriesLastTerm += 1 print check

Ans: 76576500

# Number of Inversions in an Unsorted Array: Python Code

This is my solution to the first programming assignment of Tim Roughgarden’s course on Algorithms  that was due 12:30 PM IST today. Here’s the question quoted as it is:

Programming Question-1
Download the text file here. (Right click and save link as) This file contains all of the 100,000 integers between 1 and 100,000 (inclusive) in some order, with no integer repeated.

Your task is to compute the number of inversions in the file given, where the ith row of the file indicates the ith entry of an array.
Because of the large size of this array, you should implement the fast divide-and-conquer algorithm covered in the video lectures. The numeric answer for the given input file should be typed in the space below.
So if your answer is 1198233847, then just type 1198233847 in the space provided without any space / commas / any other punctuation marks. You can make up to 5 attempts, and we’ll use the best one for grading.
(We do not require you to submit your code, so feel free to use any programming language you want — just type the final numeric answer in the following space.)

My Solution

I modified an earlier code I wrote for merge sort to arrive at the solution. It needed just a couple of modifications, and if you look carefully, it turns out that the number of inversions are unearthed each and every time we merge two sorted sub-arrays. So, intuitively, if the merge sort algorithm was O(nlog2 n), it would take almost as many operations for counting inversions. In python, the code to sort and count inversions in an array of 10,000 integers took less than 3 seconds.

 # load contents of text file into a list # numList NUMLIST_FILENAME = "IntegerArray.txt" inFile = open(NUMLIST_FILENAME, 'r') with inFile as f: numList = [int(integers.strip()) for integers in f.readlines()] count = 0 def inversionsCount(x): global count midsection = len(x) / 2 leftArray = x[:midsection] rightArray = x[midsection:] if len(x) > 1: # Divid and conquer with recursive calls # to left and right arrays similar to # merge sort algorithm inversionsCount(leftArray) inversionsCount(rightArray) # Merge sorted sub-arrays and keep # count of split inversions i, j = 0, 0 a = leftArray; b = rightArray for k in range(len(a) + len(b) + 1): if a[i] <= b[j]: x[k] = a[i] i += 1 if i == len(a) and j != len(b): while j != len(b): k +=1 x[k] = b[j] j += 1 break elif a[i] > b[j]: x[k] = b[j] count += (len(a) - i) j += 1 if j == len(b) and i != len(a): while i != len(a): k+= 1 x[k] = a[i] i += 1 break return x # call function and output number of inversions inversionsCount(numList) print count
view raw countInversions.py hosted with ❤ by GitHub

Test my solution here with your own test cases.

# Algorithmic Game Theory Lecture Videos and Notes

Link to Stanford professor, Tim Roughgarden’s video lectures on algorithmic game theory (AGT):

2013 Iteration
http://theory.stanford.edu/~tim/f13/f13.html

2014 Iteration
http://theory.stanford.edu/~tim/f14/f14.html

I’m currently doing his Coursera MOOC on algorithms, divided into 2 parts:

I’m teaching my algorithmic game theory course at Stanford this quarter, and this time around I’m posting lecture videos and notes.  The videos are a static shot of my blackboard lectures, not MOOC-style videos.

The course home page is here.  Week 1 videos and notes, covering several motivating examples and some mechanism design basics, are already available.  This week (Week 2) we’ll prove the correspondence between monotone and implementable allocation rules in single-parameter environments, and introduce algorithmic mechanism design via Knapsack auctions.

The ten-week course has roughly four weeks of lectures on mechanism design, three weeks on the inefficiency of equilibria (e.g., the price of anarchy), and three weeks on algorithms for and the complexity of learning and computing equilibria. Periodically, I’ll post updates on the course content in this space.  I would be very happy to receive comments, corrections, and criticisms on the course organization and content.

View original post

# The Merge Sort — Python Code

I have just begun working on a MOOC on algorithms offered by Stanford. Since this course gives us the liberty to choose a programming language, there isn’t any code discussed in those lectures. I plan to convert any algorithm discussed in those lectures into Python code. Since Merge Sort was the first algorithm discussed, I’m starting with that.

Merge Sort is supposedly a good introduction to divide and conquer algorithms, greatly improving upon selection, insertion and bubble sort techniques, especially when input size increases.

Pseudocode:

— Recursively sort the first half of the input array.
— Recursively sort the second half of the input array.
— Merge two sorted sub-lists into one list.

C = output [length = n]
A = 1st sorted array [n/2]
B = 2nd sorted array [n/2]
i = 0 or 1 (depending on the programming language)
j = 0 or 1 (depending on the programming language)

for k = 1 to n

if A(i) < B(j)
C(k) = A(i)
i = i + 1

else if A(i) > B(j)
C(k) = B(j)
j = j + 1

Note: the pseudocode for the merge operation ignores the end cases.

Visualizing the algorithm can be done in 2 stages — first, the recursive splitting of the arrays, 2 each 2 at a time, and second, the merge operation.

Here’s the Python code to merge sort an array.

 # Code for the merge subroutine def merge(a,b): """ Function to merge two arrays """ c = [] while len(a) != 0 and len(b) != 0: if a[0] < b[0]: c.append(a[0]) a.remove(a[0]) else: c.append(b[0]) b.remove(b[0]) if len(a) == 0: c += b else: c += a return c # Code for merge sort def mergesort(x): """ Function to sort an array using merge sort algorithm """ if len(x) == 0 or len(x) == 1: return x else: middle = len(x)/2 a = mergesort(x[:middle]) b = mergesort(x[middle:]) return merge(a,b)
view raw mergesort.py hosted with ❤ by GitHub

We can divide a list in half log2 n times where n is the length of the list. The second process is the merge. Each item in the list will eventually be processed and placed on the sorted list. So the merge operation which results in a list of size n requires n operations. The result of this analysis is that log2 n splits, each of which costs n for a total of nlog2 n operations.

Other Algorithms:
Karatsuba Integer Multiplication Algorithm
Quick Sort Python Code